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\title{{\Huge Homework3\\ 数值分析}}
\author{{\LARGE 3190105815 信息与计算科学\ 行一凡}}
\date{{\LARGE \today}}

\begin{document}
	\maketitle
	\section*{Exercise I}
	由题可得插值条件$ p(0) = 0,\ p(1) = 1,\ p^{\prime}(1) = -3,\ p^{\prime\prime}(1) = 6 $，则可计算差分表
	\begin{center}
		\begin{tabular}{c|cccc}
			$ 0 $ & 0 & & &\\ 
			$ 1 $ & 1 & 1 & &\\ 
			$ 1 $ & 1 & -3 & -4 &\\ 
			$ 1 $ & 1 & -3 & 3 & 7 
		\end{tabular}
	\end{center}
	故有$ p(x) = 7x^3 - 18x^2 + 12x $，而$ p^{\prime\prime}(0) = -36 \neq0 $，则$ s(x) $不是自然样条。
	
	\section*{Exercise II}
	\begin{enumerate}[(a).]
		\item 在区间$ [x_1, x_2] $上，若已知$ f(x_1),\ f(x_{2}),\ f^{\prime}(x_{2}) $，则应用 Hermite 插值可得到唯一的$ p(x) $满足$ p\in \mathbb{P}_2 $，于是通过任取$ f^{\prime}(x_{2}) $能计算出不同的结果，最终得到的$ s(x) $也是不同的，因此需要更多条件约束。
		\item 在$ [x_i, x_{i+1}] $上有$ p_i(x_i) = f_i,\ p_i^{\prime}(x_i) = f_i,\ p_i(x_{i+1}) = f_{i+1} $，计算差分表得
		\begin{center}
			\begin{tabular}{c|ccc}
				$ x_i $ & $ f_i $ & &\\ 
				$ x_i $ & $ f_i $ & $ m_i $ &\\ 
				$ x_{i+1} $ & $ f_{i+1} $ & $ \frac{f_{i+1}-f_i}{x_{i+1}-x_i} $ & $ \frac{f_{i+1}-f_i}{(x_{i+1}-x_i)^2} - \frac{m_i}{x_{i+1}-x_i} $
			\end{tabular}
		\end{center}
		从而有
		\begin{equation}\label{key}
			p_i(x) = f_i + m_i(x-x_i) + \left[\dfrac{f_{i+1}-f_i}{(x_{i+1}-x_i)^2} - \dfrac{m_i}{x_{i+1}-x_i}\right](x-x_i)^2
		\end{equation}
		\item 由于$ p_i^{\prime}(x_i) = p_{i-1}^{\prime}(x_i) $，代入解得
		\begin{equation}\label{key}
			\begin{aligned}
				m_i &= m_{i-1} + \left[\dfrac{f_{i}-f_{i-1}}{(x_{i}-x_{i-1})^2} - \dfrac{m_{i-1}}{x_{i}-x_{i-1}}\right]\cdot 2(x_i-x_{i-1}) \\ 
				&= f[x_{i-1},x_i]-m_{i-1} 
			\end{aligned}
		\end{equation}
		且已知$ m_1 = f^{\prime}(a) $，从而可以按上式递推计算$ m_2,m_3,\cdots,m_{n-1} $.
	\end{enumerate}

	\section*{Exercise III}
	已知插值条件$ s_2(0) = 1+c,\ s_2^{\prime}(0) = s_1^{\prime}(0) = 3c,\ s_2^{\prime\prime}(0) = s_1^{\prime\prime}(0) = 6c $，可得差分表
	\begin{center}
		\begin{tabular}{c|cccc}
			$ 0 $ & $ 1+c $ & & &\\ 
			$ 0 $ & $ 1+c $ & $ 3c $ & &\\ 
			$ 0 $ & $ 1+c $ & $ 3c $ & $ 3c $ &\\ 
			$ 1 $ & $ s_2(1) $ & $ s_2(1)-c-1 $ & $ s_2(1)-4c-1 $ & $ s_2(1)-7c-1 $ 
		\end{tabular}
	\end{center}
	即有
	\begin{equation}\label{key}
		s_2(x) = 1+c+3cx+3cx^2+(s_2(1)-7c-1)x^3
	\end{equation}
	由于$ s(x) $是自然样条，故$ s_2^{\prime\prime}(1) = s^{\prime\prime}(1) = 0 $，代入解得$ s_2(1) = 6c+1 $，从而
	\begin{equation}\label{key}
		s_2(x) = 1+c+3cx+3cx^2-cx^3
	\end{equation}
	若要$ s(1) = -1 $，则$ s_2(1) = -1 $，解得$ c = -1 $.
	
	\section*{Exercise IV}
	\begin{enumerate}[(a).]
		\item 已知$ f(-1) = 0,\ f(0) = 1,\ f(1) = 0 $，不妨设$ x_1 = -1,\ x_2 = 0,\ x_3 = 1 $，令$ M_i = f^{\prime\prime}(x_i) $，则由 Lemma 4.3 可得
		\begin{equation}\label{key}
			\mu_2 = \dfrac{x_2-x_1}{x_3-x_1} = \dfrac{1}{2},\ \lambda_2 = \dfrac{x_3-x_2}{x_3-x_1} = \dfrac{1}{2}
		\end{equation}
		由 Lemma 4.5 可得
		$ \frac{1}{2}M_1 + 2M_2 + \frac{1}{2}M_3 = 6f[x_1, x_2, x_3] $
		计算差分表得到
		\begin{center}
			\begin{tabular}{c|ccc}
				$ -1 $ & 0 & & \\ 
				$ 0 $ & 1 & 1 & \\ 
				$ 1 $ & 0 & -1 & -1 
			\end{tabular}
		\end{center}
		由$ s(x) $是自然样条，故$ M_1 = M_3 = 0 $，则有
		\begin{equation}\label{key}
			M_2 = 3f[x_1, x_2, x_3] = -3
		\end{equation}
		令$ m_i = f^{\prime}(x_i) $，则在$ [x_1,x_2] $上计算差分表得
		\begin{center}
			\begin{tabular}{c|cccc}
				$ -1 $ & 0 & & &\\ 
				$ 0 $ & 1 & 1 & &\\ 
				$ 0 $ & 1 & $ m_2 $ & $ m_2-1 $&\\ 
				$ 0 $ & 1 & $ m_2 $ & $ -\frac{3}{2} $ & $ -\frac{1}{2}-m_2 $
			\end{tabular}
		\end{center}
		从而有
		\begin{equation}\label{key}
			s_1(x) = (x+1) - (m_2-1)x(x+1) - (\dfrac{1}{2}+m_2)x^2(x+1)
		\end{equation}
		由$ s(x) $为自然样条，有$ s_i^{\prime\prime}(-1) = 0 $，而
		\begin{equation}\label{key}
			s_i^{\prime\prime}(x) = 2(m_2-1)-(\dfrac{1}{2}+m_2)(6x+2)
		\end{equation}
		代入可解得$ m_2 = 0 $，故可得
		\begin{equation}\label{key}
			s_1(x) = (x+1) - x(x+1) - \dfrac{1}{2}x^2(x+1) = 1 - \dfrac{3}{2}x^2 - \dfrac{1}{2}x^3
		\end{equation}
		同理，在$ [x_2,x_3] $上计算差分表有
		\begin{center}
			\begin{tabular}{c|cccc}
				$ 0 $ & 1 & & &\\ 
				$ 0 $ & 1 & 0 & &\\ 
				$ 0 $ & 1 & $ 0 $ & $ -\frac{3}{2} $&\\ 
				$ 1 $ & 0 & $ -1 $ & $ -1 $ & $ \frac{1}{2} $
			\end{tabular}
		\end{center}
		则有
		\begin{equation}\label{key}
			s_2(x) = 1-\dfrac{3}{2}x^2+\dfrac{1}{2}x^3
		\end{equation}
		综上有
		\begin{equation}\label{key}
			s(x) = \left\{
			\begin{aligned}
				&1 - \dfrac{3}{2}x^2 - \dfrac{1}{2}x^3,\ x\in[-1,0]\\ 
				&1-\dfrac{3}{2}x^2+\dfrac{1}{2}x^3,\ x\in[0,1]
			\end{aligned}
			\right.
		\end{equation}
	
		\item 直接计算求二阶导有
		\begin{equation}\label{key}
			s^{\prime\prime}(x) = \left\{
			\begin{aligned}
				&-3-3x,\ x\in[-1,0]\\ 
				&-3+3x,\ x\in[0,1]
			\end{aligned}
			\right.
		\end{equation}
		\begin{equation}\label{key}
			\int_{-1}^{1}[s^{\prime\prime}(x)]^2dx = \int_{-1}^{0}9(x+1)^2dx + \int_{0}^{1}9(x-1)^2dx = 6
		\end{equation}
		\begin{enumerate}[(i).]
			\item 由之前计算的差分表，可得在$ x = -1,0,1 $上的插值多项式为$ g(x) = 1-x^2 $，故有
			\begin{equation}\label{key}
				\begin{aligned}
					\int_{-1}^{1}[g^{\prime\prime}(x)]^2dx &= \int_{-1}^{1}[-2]^2dx = 8\\ 
					&> \int_{-1}^{1}[s^{\prime\prime}(x)]^2dx
				\end{aligned}
			\end{equation}
		
			\item $ g(x) = f(x) = \cos(\frac{\pi}{2}x),\ g^{\prime\prime}(x) = -\frac{\pi^2}{4}\cos(\frac{\pi}{2}x) $，则
			\begin{equation}\label{key}
					\begin{aligned}
					\int_{-1}^{1}[g^{\prime\prime}(x)]^2dx &= \int_{-1}^{1}\left[\dfrac{\pi^2}{4}\cos\left(\dfrac{\pi}{2}x\right)\right]^2dx\\ 
					&= \int_{-1}^{1}\dfrac{\pi^4}{32}(1+\cos(\pi x))dx\\ 
					&= \dfrac{\pi^4}{16}\\ 
					&\approx 6.088 \\ 
					&> \int_{-1}^{1}[s^{\prime\prime}(x)]^2dx
				\end{aligned}
			\end{equation}
		\end{enumerate}
	\end{enumerate}

	\section*{Exercise V}
	\begin{enumerate}[(a).]
		\item 由递归定义
		\begin{equation}\label{key}
			B_i^1(x) = \left\{
				\begin{aligned}
					&\dfrac{x-t_{i-1}}{t_{i}-t_{i-1}},\ x\in[t_{i-1},t_i]\\ 
					&\dfrac{t_{i+1}-x}{t_{i+1}-t_{i}},\ x\in[t_{i},t_{i+1}]
				\end{aligned}
			\right.
		\end{equation}
		\begin{equation}\label{key}
			B_i^2(x) = \dfrac{x-t_{i-1}}{t_{i+1}-t_{i-1}}B_i^1(x) + \dfrac{t_{i+2}-x}{t_{i+2}-t_{i}}B_{i+1}^1(x)
		\end{equation}
		则对$ x\in[t_{i-1},t_i] $，有
		\begin{equation}\label{key}
			B_i^2(x) = \dfrac{(x-t_{i-1})^2}{(t_{i+1}-t_{i-1})(t_{i}-t_{i-1})}
		\end{equation}
		对$ x\in[t_i,t_{i+1}] $，有
		\begin{equation}\label{key}
			B_i^2(x) = \dfrac{(x-t_{i-1})(t_{i+1}-x)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})} + \dfrac{(t_{i+2}-x)(x-t_{i})}{(t_{i+2}-t_{i})(t_{i+1}-t_i)}
		\end{equation}
		对$ x\in[t_{i+1},t_{i+2}] $，有
		\begin{equation}\label{key}
			B_i^2(x) = \dfrac{(t_{i+2}-x)^2}{(t_{i+2}-t_{i})(t_{i+2}-t_{i+1})}
		\end{equation}
		对于$ x $的其它情况，$ B_i^2(x) = 0 $.
		
		\item 直接求导得
		\begin{equation}\label{key}
			\dfrac{d}{dx}B_i^2(x) = \left\{
				\begin{aligned}
					&\frac{2(x-t_{i-1})}{(t_{i+1}-t_{i-1})(t_{i}-t_{i-1})},\ &x\in[t_{i-1},t_i]\\ 
					&\frac{t_{i-1}+t_{i+1}-2x}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})} + \frac{t_i+t_{i+2}-2x}{(t_{i+2}-t_{i})(t_{i+1}-t_i)},\ &x\in[t_i,t_{i+1}]\\ 
					&\frac{2(x-t_{i+2})}{(t_{i+2}-t_{i})(t_{i+2}-t_{i+1})},\ &x\in[t_{i+1},t_{i+2}]\\ 
					&0,\ &\mathrm{otherwise}
				\end{aligned}
				\right.
		\end{equation}
		代入$ t_i,t_{i+1} $得到
		\begin{equation}\label{key}
			\dfrac{d}{dx}B_i^2(t_i-0) = \dfrac{2}{t_{i+1}-t_{i-1}},\ \dfrac{d}{dx}B_i^2(t_{i+1}+0) = -\dfrac{2}{t_{i+2}-t_{i}}
		\end{equation}
	
		\begin{equation}\label{key}
			\begin{aligned}
				\dfrac{d}{dx}B_i^2(t_i+0) &= \dfrac{t_{i-1}-t_i+t_{i+1}-t_i}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}+\dfrac{1}{t_{i+1}-t_i}\\ 
				&= \dfrac{t_{i-1}-t_i+t_{i+1}-t_i+t_{i+1}-t_{i-1}}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}\\ 
				&= \dfrac{2}{t_{i+1}-t_{i-1}}
			\end{aligned}
		\end{equation}
		
		\begin{equation}\label{key}
			\begin{aligned}
				\dfrac{d}{dx}B_i^2(t_{i+1}-0) &= -\dfrac{1}{t_{i+1}-t_{i}}+\dfrac{t_i-t_{i+1}+t_{i+2}-t_{i+1}}{(t_{i+2}-t_i)(t_{i+1}-t_i)}\\ 
				&= \dfrac{-t_{i+2}+t_{i}+t_i-t_{i+1}+t_{i+2}-t_{i+1}}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})}\\ 
				&= -\dfrac{2}{t_{i+2}-t_{i}}
			\end{aligned}
		\end{equation}
		从而$ \frac{d}{dx}B_i^2(x) $在$ t_i,t_{i+1} $连续。
		
		\item 由$ (b) $有$ \frac{d}{dx}B_i^2(t_{i-1}) = 0,\ \frac{d}{dx}B_i^2(t_{i+1}) = -\frac{2}{t_{i+2}-t_{i}} < 0 $，且$ \frac{d}{dx}B_i^2(x) $在$ [t_{i-1},t_i] $上严格单增，在$ [t_i,t_{i+1}] $上严格单减，因此$ \frac{d}{dx}B_i^2(t_{i}) > 0 $，则$ \frac{d}{dx}B_i^2(x) $在$ (t_{i-1},t_{i+1}) $上有唯一零点$ x^*\in(t_{i},t_{i+1}) $，计算得
		\begin{equation}\label{key}
			k = -\dfrac{2}{(t_{i+1}-t_{i-1})(t_{i+1}-t_i)}-\dfrac{2}{(t_{i+2}-t_i)(t_{i+1}-t_i)}
		\end{equation}
		\begin{equation}\label{key}
			x^* = t_{i+1}+\dfrac{2}{k(t_{i+2}-t_i)} = \dfrac{t_{i+1}t_{i+2}-t_it_{i-1}}{t_{i+2}-t_i+t_{i+1}-t_{i-1}}
		\end{equation}
	
		\item 由于在$ [t_{i+1},t_{i+2}] $上$ \frac{d}{dx}B_i^2(x)\le 0 $，又由$ (c) $可得$ B_i^2(x) $在$ x^* $处取得极大值，代入得
		\begin{equation}\label{key}
			\begin{aligned}
				B_i^2(x^*) &= \dfrac{(x^*-t_{i-1})(t_{i+1}-x^*)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})} + \dfrac{(t_{i+2}-x^*)(x^*-t_{i})}{(t_{i+2}-t_{i})(t_{i+1}-t_i)}\\ 
				&= \dfrac{(t_{i+2}-t_{i-1})(t_{i+1}-t_i)(t_{i+1}-t_{i-1})^2}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})(t_{i+2}-t_i+t_{i+1}-t_{i-1})^2}\\  &+ \dfrac{(t_{i+2}-t_{i-1})(t_{i+1}-t_i)(t_{i+2}-t_i)^2}{(t_{i+2}-t_{i})(t_{i+1}-t_i)(t_{i+2}-t_i+t_{i+1}-t_{i-1})^2}\\ 
				&= \dfrac{(t_{i+2}-t_{i-1})(t_{i+1}-t_{i-1})}{(t_{i+2}-t_i+t_{i+1}-t_{i-1})^2} + \dfrac{(t_{i+2}-t_{i-1})(t_{i+2}-t_i)}{(t_{i+2}-t_i+t_{i+1}-t_{i-1})^2}\\ 
				&= \dfrac{t_{i+2} - t_{i-1}}{t_{i+2}-t_i+t_{i+1}-t_{i-1}} < 1
			\end{aligned}
		\end{equation}
		而$ B_i^2(t_{i-1}) = 0 $，且由$ (c),(d) $中对其在不同区间单调性的分析，有$ B_i^2(x)\in\left[0,1\right) $.
		
		\item 如图所示
		\begin{figure}[!htb]			% 嵌套figure环境
			\centering			% 图片居中
			\includegraphics[width=0.5\textwidth]{B_i^2.png}
		\end{figure}
	\end{enumerate}

	\section*{Exercise VI}
	根据
	\begin{equation}\label{key}
		(t-x)_+^2 = \left\{
		\begin{aligned}
			&(t-x)^2,\ x\le t\\ 
			&0,\ x>t
		\end{aligned}
		\right.
	\end{equation}
	从而有
	\begin{equation}\label{key}
		[t_{i-1},t_i](t-x)_+^2 = \dfrac{(t_i-x)_+^2-(t_{i-1}-x)_+^2}{t_i-t_{i-1}}
	\end{equation}
	\begin{equation}\label{key}
		\begin{aligned}
			[t_{i-1},t_i,t_{i+1}](t-x)_+^2 &= \dfrac{[t_{i+1},t_i](t-x)_+^2-[t_{i-1},t_i](t-x)_+^2}{t_{i+1}-t_{i-1}}\\ 
			&= \dfrac{1}{t_{i+1}-t_{i-1}}\left[\dfrac{(t_{i+1}-x)_+^2-(t_{i}-x)_+^2}{t_{i+1}-t_{i}} - \dfrac{(t_i-x)_+^2-(t_{i-1}-x)_+^2}{t_i-t_{i-1}}\right]
		\end{aligned}
	\end{equation}
	\begin{equation}\label{key}
		\begin{aligned}
			&\indent(t_{i+2}-t_{i-1})[t_{i-1},t_i,t_{i+1},t_{i+2}](t-x)_+^2\\ &= [t_{i},t_{i+1},t_{i+2}](t-x)_+^2 - [t_{i-1},t_i,t_{i+1}](t-x)_+^2\\ 
			&= \dfrac{1}{t_{i+2}-t_{i}}\left[\dfrac{(t_{i+2}-x)_+^2-(t_{i+1}-x)_+^2}{t_{i+2}-t_{i+1}} - \dfrac{(t_{i+1}-x)_+^2-(t_{i}-x)_+^2}{t_{i+1}-t_{i}}\right]\\ 
			&\ \ \ - \dfrac{1}{t_{i+1}-t_{i-1}}\left[\dfrac{(t_{i+1}-x)_+^2-(t_{i}-x)_+^2}{t_{i+1}-t_{i}} - \dfrac{(t_i-x)_+^2-(t_{i-1}-x)_+^2}{t_i-t_{i-1}}\right]
		\end{aligned}
	\end{equation}
	代入$ (t-x)_+^2 $的表达式即证。
	
	\section*{Exercise VII}
	由 Theorem 4.35 可得
	\begin{equation}\label{key}
		\dfrac{d}{dx}B_i^{n+1}(x) = \dfrac{(n+1)B_i^{n}(x)}{t_{i+n}-t_{i-1}}-\dfrac{(n+1)B_{i+1}^{n}(x)}{t_{i+n+1}-t_i}
	\end{equation}
	两边同时在$ [t_{i-1},t_{i+n+1}] $上积分可得
	\begin{equation}\label{key}
		\begin{aligned}
			\int_{t_{i-1}}^{t_{i+n+1}}\dfrac{d}{dx}B_i^{n+1}(x)dx &= \int_{t_{i-1}}^{t_{i+n+1}}\left[\dfrac{(n+1)B_i^{n}(x)}{t_{i+n}-t_{i-1}}-\dfrac{(n+1)B_{i+1}^{n}(x)}{t_{i+n+1}-t_i}\right]dx\\ 
			&= (n+1)\left[\dfrac{1}{{t_{i+n}-t_{i-1}}}\int_{t_{i-1}}^{t_{i+n}}B_i^{n}(x)dx-\dfrac{1}{t_{i+n+1}-t_i}\int_{t_{i}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx\right]
		\end{aligned}
	\end{equation}
	由于$ \int_{t_{i-1}}^{t_{i+n+1}}\frac{d}{dx}B_i^{n+1}(x)dx = B_i^{n+1}(t_{i+n+1})-B_i^{n+1}(t_{i-1}) = 0  $，故有
	\begin{equation}\label{key}
		\dfrac{1}{{t_{i+n}-t_{i-1}}}\int_{t_{i-1}}^{t_{i+n}}B_i^{n}(x)dx = \dfrac{1}{t_{i+n+1}-t_i}\int_{t_{i}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx
	\end{equation}
	从而在支集区间上的积分与$ i $无关。
	
	\section*{Exercise VIII}
	\begin{enumerate}[(a).]
		\item 对$ m=4,n=2 $有
		\begin{equation}\label{key}
			\begin{aligned}
				[x_i,x_{i+1},x_{i+2}]x^4 &= \dfrac{[x_{i+1},x_{i+2}]x^4-[x_i,x_{i+1}]x^4}{x_{i+2}-x_i}\\ 
				&= \dfrac{\frac{x_{i+2}^4-x_{i+1}^4}{x_{i+2}-x_{i+1}}-\frac{x_{i+1}^4-x_i^4}{x_{i+1}-x_i}}{x_{i+2}-x_i}\\ 
				&= \dfrac{(x_{i+2}^3+x_{i+2}^2x_{i+1}+x_{i+2}x_{i+1}^2+x_{i+1}^3)-(x_{i+1}^3+x_{i+1}^2x_i+x_{i+1}x_i^2+x_{i}^3)}{x_{i+2}-x_i}\\ 
				&= \dfrac{(x_{i+2}^3-x_{i}^3)+(x_{i+2}^2-x_i^2)x_{i+1}+(x_{i+2}-x_i)x_{i+1}^2}{x_{i+2}-x_i}\\ 
				&= (x_{i+2}^2+x_{i+2}x_i+x_i^2)+(x_{i+2}+x_i)x_{i+1}+x_{i+1}^2\\ 
				&= x_{i+2}^2 + x_{i+1}^2 + x_i^2 + x_ix_{i+1} + x_{i+1}x_{i+2} + x_{i+2}x_{i}
			\end{aligned}
		\end{equation}
		而由完全对称多项式的定义可得
		\begin{equation}\label{key}
			\begin{aligned}
				\tau_{2}(x_i,x_{i+1},x_{i+2}) &= \sum_{i\le j_1\le j_2\le i+2}x_{j_1}x_{j_2}\\ 
				&= x_{i+2}^2 + x_{i+1}^2 + x_i^2 + x_ix_{i+1} + x_{i+1}x_{i+2} + x_{i+2}x_{i}
			\end{aligned}
		\end{equation}
		则$ \tau_{2}(x_i,x_{i+1},x_{i+2}) = [x_i,x_{i+1},x_{i+2}]x^4 $.
		
		\item 由 Lemma 4.46 有
		\begin{equation}\label{key}
			\begin{aligned}
				&\ \ \ (x_{n+1}-x_n)\tau_k(x_1,\cdots,x_n,x_{n+1})\\ 
				&= \tau_{k+1}(x_1,\cdots,x_n,x_{n+1})-\tau_{k+1}(x_1,\cdots,x_n)\\ 
				&\ \ \ -x_n\tau_k(x_1,\cdots,x_n,x_{n+1})\\ 
				&= \tau_{k+1}(x_2,\cdots,x_n,x_{n+1})+x_n\tau_k(x_1,\cdots,x_n,x_{n+1})\\ 
				&\ \ \ -\tau_{k+1}(x_1,\cdots,x_n)-x_n\tau_k(x_1,\cdots,x_n,x_{n+1})\\ 
				&= \tau_{k+1}(x_2,\cdots,x_n,x_{n+1})-\tau_{k+1}(x_1,\cdots,x_n)
			\end{aligned}
		\end{equation}
		$ n=0 $时，有$ \tau_m(x_i) = [x_i]x^m $成立，则假设对$ n = k-1 $成立，则有
		\begin{equation}\label{key}
			\begin{aligned}
				&\ \ \ \tau_{m-k}(x_i,\cdots,x_{i+k})\\ 
				&= \dfrac{\tau_{m-k+1}(x_{i+1},\cdots,x_{i+k})-\tau_{m-k+1}(x_{i},\cdots,x_{i+k-1})}{x_{i+k}-x_{i}}\\ 
				&= \dfrac{[x_{i+1},\cdots,x_{i+k}]x^{m}-[x_{i},\cdots,x_{i+k-1}]x^{m}}{x_{i+k}-x_{i}}\\ 
				&= [x_{i},\cdots,x_{i+k}]x^m
			\end{aligned}
		\end{equation}
		从而由归纳法即证。
	\end{enumerate}
	
\end{document}